Question

Find the equation of a straight line through the intersection of lines 5x - 6y = 2,
3x + 2y = 10 and perpendicular to the line 4x - 7y +13 = 0
​

Answer 1

$given \: lines \: 5x - 6y = 2 \: and \: 3x + 2y = 10$

$5x - 6y - 2 = 0 ..........(i)$

$3x + 2y - 10 = 0..........(ii)$

Doing, (i)×1+(ii)×3

we get, x=−1

Replacing value of x in the equations we have, y=−1

∴ the lines intesect at (−1,−1)

The slope of the line 3x−5y+11=0 is $\frac{3}{5}$

So, the slope of a line perpendicular to it is

$\frac{ - 5}{3}$

Hence, the equation of the required line is

$y - ( - 1) = \frac{ - 5}{3} (x - ( - 1))$

$= > y + 1 = \frac{ - 5}{3} (x + 1)$

$= > 5x + 3y + 8 = 0$

HOPE IT'S CLARIFIES YOU ❣️❣️

HAVE A GREAT DAY AHEAD ❣️❣️